Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

On a divergent integral

  1. Mar 30, 2012 #1
    greetings . we have the integral :
    [tex] \lim_{T\to \infty }\int_{2-iT}^{2+iT}\frac{(s-1)^{n}}{s}ds[/tex]

    which diverges for every value of n except [itex] n=0 [/itex]
    if we perform the change of variables :

    [tex] s\rightarrow \frac{1}{s}[/tex]

    then :
    [tex]\lim_{T\to \infty }\int_{2-iT}^{2+iT}\frac{(s-1)^{n}}{s}ds=\int_{-i}^{i}\frac{(1-s)^{n}}{s^{n+1}}ds[/tex]

    which converges . am i missing something here , or is this correct !?
     
  2. jcsd
  3. Mar 30, 2012 #2
    Can you solve it for n=2 using antiderivatives? That is, what is:

    [tex]\lim_{T\to\infty}\left(-2s+\frac{s^2}{2}+\log(s)\right)_{2-iT}^{2+iT}[/tex]
     
  4. Mar 30, 2012 #3

    mathman

    User Avatar
    Science Advisor
    Gold Member

    You need to split the original integration range into two parts at s = 0. Now when you change s to 1/s, you will be able to get the correct integration limits. Note also that the original T becomes 1/T.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: On a divergent integral
  1. Divergent integrals. (Replies: 2)

  2. Diverging Integral (Replies: 1)

Loading...